Question 14

Analyzing the hybridization and shape of each species using VSEPR theory:

1. $SOCl_2$: $S$ has 6 valence electrons. 1 double bond with $O$, 2 single bonds with $Cl$, and 1 lone pair. Steric Number = 4 ($sp^3$). Shape: Pyramidal.
2. $XeOF_4$: $Xe$ has 8 valence electrons. 1 double bond with $O$, 4 single bonds with $F$, and 1 lone pair. Steric Number = 6 ($sp^3d^2$). Shape: Square Pyramidal.
3. $ClF_3$: $Cl$ has 7 valence electrons. 3 single bonds with $F$, and 2 lone pairs. Steric Number = 5 ($sp^3d$). Shape: T-shaped.
4. $ClF_5$: $Cl$ has 7 valence electrons. 5 single bonds with $F$, and 1 lone pair. Steric Number = 6 ($sp^3d^2$). Shape: Square Pyramidal.
5. $XeF_5^+$: $Xe^+$ has 7 valence electrons. 5 single bonds with $F$, and 1 lone pair. Steric Number = 6 ($sp^3d^2$). Shape: Square Pyramidal.
6. $SO_3^{2-}$: $S$ has 6 valence electrons. 3 sigma bonds (with negative charges on oxygens) and 1 lone pair. Steric Number = 4 ($sp^3$). Shape: Pyramidal.
7. $XeF_3^+$: $Xe^+$ has 7 valence electrons. 3 single bonds with $F$, and 2 lone pairs. Steric Number = 5 ($sp^3d$). Shape: T-shaped.
8. $SF_4$: $S$ has 6 valence electrons. 4 single bonds with $F$, and 1 lone pair. Steric Number = 5 ($sp^3d$). Shape: See-saw.

Matching with List-I: (P) See-saw: Only $SF_4$ (Total = 1). So P $\rightarrow$ 1. (Q) T-Shaped: $ClF_3$ and $XeF_3^+$ (Total = 2). So Q $\rightarrow$ 2. (R) Trigonal Planar: None (Total = 0). So R $\rightarrow$ 5. (S) Square Pyramidal: $XeOF_4$, $ClF_5$, and $XeF_5^+$ (Total = 3). So S $\rightarrow$ 3.

Correct Option: (A)